Find the Domain Regions of y=-4x²+6: Complete Function Analysis

To solve the problem, we will find where the quadratic function $y = -4x^2 + 6$ is equal to zero.

Set the equation to zero to find the roots:

$-4x^2 + 6 = 0$

$-4x^2 = -6$

$x^2 = \frac{6}{4}$

$x^2 = \frac{3}{2}$

Take the square root of both sides:

$x = \pm \sqrt{\frac{3}{2}}$

$x = \pm \frac{\sqrt{6}}{2}$ (since $\sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$)

Now identify the intervals:

• Interval 1: $x < -\frac{\sqrt{6}}{2}$
• Interval 2: $-\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2}$
• Interval 3: $x > \frac{\sqrt{6}}{2}$

Test each interval to determine positivity or negativity:

• For Interval 1 ($x < -\frac{\sqrt{6}}{2}$): The parabola opens downwards and is negative outside roots.
• For Interval 2 ($-\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2}$): This interval is between the roots, so $y > 0$.
• For Interval 3 ($x > \frac{\sqrt{6}}{2}$): Again, as the parabola opens downward, $y < 0$.

Therefore, the positive domain is $-\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2}$, and the negative domains are $x < -\frac{\sqrt{6}}{2}$ and $x > \frac{\sqrt{6}}{2}$.

The correct answer is choice 4.

$x > 0: -\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2}$

$x > \frac{\sqrt{6}}{2}$ or $x < 0: x < -\frac{\sqrt{6}}{2}$